
Let a be a vector whose components are called [a_{1}, a_{2}, a_{3}], and let b and c be vectors as well.
The cross product a�b = [a_{2}b_{3}a_{3}b_{2}, a_{3}b_{1}a_{1}b_{3}, a_{1}b_{2}a_{2}b_{1}]. The cross product is perpendicular to both a and b, and has a magnitude equal to a b sin θ, which is the area of a parallelogram, two sides of which are formed by vectors a and b.
The dot product a·b = a b cos θ, and in three dimensions, a·b = a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3}
The triple product (a�b)·c is a scalar representing the "signed volume" of a parallelepiped.
a�(b�c) = (c�a)b  (b�a)c is Lagrange's formula, sometimes called the "CAB Minus BAC" identity
Proof:
Let a = [a_{1},a_{2},a_{3}], and let b and c be defined similarly.
b�c = [b_{2}c_{3}b_{3}c_{2}, b_{3}c_{1}b_{1}c_{3}, b_{1}c_{2}b_{2}c_{1}]
a�(b�c) = [a_{2}(b_{1}c_{2}b_{2}c_{1})a_{3}(b_{3}c_{1}b_{1}c_{3}), a_{3}(b_{2}c_{3}b_{3}c_{2})a_{1}(b_{1}c_{2}b_{2}c_{1}), a_{1}(b_{3}c_{1}b_{1}c_{3})a_{2}(b_{2}c_{3}b_{3}c_{2})]
a�(b�c) = [a_{2}b_{1}c_{2}a_{2}b_{2}c_{1}a_{3}b_{3}c_{1}+a_{3}b_{1}c_{3}, a_{3}b_{2}c_{3} a_{3}b_{3}c_{2}a_{1}b_{1}c_{2}+a_{1}b_{2}c_{1}, a_{1}b_{3}c_{1} a_{1}b_{1}c_{3}a_{2}b_{2}c_{3}+a_{2}b_{3}c_{2}](c�a)b = [(a_{1}c_{1}+a_{2}c_{2}+a_{3}c_{3})b_{1}, (a_{1}c_{1}+a_{2}c_{2}+a_{3}c_{3})b_{2}, (a_{1}c_{1}+a_{2}c_{2}+a_{3}c_{3})b_{3}]
(c�a)b = [a_{1}b_{1}c_{1}+a_{2}b_{1}c_{2}+a_{3}b_{1}c_{3}, a_{1}b_{2}c_{1}+a_{2}b_{2}c_{2}+a_{3}b_{2}c_{3}, a_{1}b_{3}c_{1}+a_{2}b_{3}c_{2}+a_{3}b_{3}c_{3}](b�a)c = [(a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})c_{1}, (a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})c_{2}, (a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})c_{3}]
(b�a)c = [a_{1}b_{1}c_{1}+a_{2}b_{2}c_{1}+a_{3}b_{3}c_{1}, a_{1}b_{1}c_{2}+a_{2}b_{2}c_{2}+a_{3}b_{3}c_{2}, a_{1}b_{1}c_{3}+a_{2}b_{2}c_{3}+a_{3}b_{3}c_{3}](c�a)b  (b�a)c = [a_{2}b_{1}c_{2}a_{2}b_{2}c_{1}+a_{3}b_{1}c_{3}a_{3}b_{3}c_{1}, a_{1}b_{2}c_{1}a_{1}b_{1}c_{2}+a_{3}b_{2}c_{3}a_{3}b_{3}c_{2}, a_{1}b_{3}c_{1}a_{1}b_{1}c_{3}+a_{2}b_{3}c_{2}a_{2}b_{2}c_{3}]
From the expansion, we see that a�(b�c) is identical to (c�a)b  (b�a)c
Some advice:
It's hard to remember what dots with what in this formula, so here's a tip. Think about the physical representation of b�c  it's a vector that's normal (that means perpendicular) to the plane defined by vectors b and c. Think of a flagpole sticking straight out of a vast bc plane. When you cross any other vector with that flagpole, the result is perpendicular to the flagpole; in other words the result is in that vast bc plane. So the vector that results from a�(b�c) is in the bc plane, which means it's a linear combination of b and c.
Now look at the formula again. a�(b�c) = (c�a)b  (b�a)c  Do you see how this formula reminds you that a�(b�c) is a linear combination of b and c? Now you need to remember just a few more things, and they should be fairly intuitive:
 The scalar factor for b is the dot product of the other two vectors, and likewise for c.
 This identity has a subtraction in it, not an addition. That should be obvious if you consider what if b=c? Clearly b�c = 0, so a�(b�c) has to be zero, too.
 Use the righthand rule to figure out whether to subtract (c�a)b  (b�a)c or vice versa.
If you follow that advice, you should be able to discern the very similar formula for (a�b)�c  then read on in this page to see if you were right.
a�(b�c) + b�(c�a) + c�(a�b) = 0
Proof:
Add the "CAB Minus BAC" identities:
a�(b�c) = (c�a)b  (b�a)c
b�(c�a) = (a�b)c  (c�b)a
c�(a�b) = (b�c)a  (a�c)b
What about (a�b)�c ?
(a�b)�c = (c�a)b  (b�c)a
Proof:
Let a = [a_{1},a_{2},a_{3}], and let b and c be defined similarly.
a�b = [a_{2}b_{3}a_{3}b_{2}, a_{3}b_{1}a_{1}b_{3}, a_{1}b_{2}a_{2}b_{1}]
(a�b)�c = [(a_{3}b_{1}a_{1}b_{3})c_{3}(a_{1}b_{2}a_{2}b_{1})c_{2}, (a_{1}b_{2}a_{2}b_{1})c_{1}(a_{2}b_{3}a_{3}b_{2})c_{3}, (a_{2}b_{3}a_{3}b_{2})c_{2}(a_{3}b_{1}a_{1}b_{3})c_{1}]
(a�b)�c = [a_{3}b_{1}c_{3}a_{1}b_{3}c_{3}a_{1}b_{2}c_{2}+a_{2}b_{1}c_{2}, a_{1}b_{2}c_{1}a_{2}b_{1}c_{1}a_{2}b_{3}c_{3}+a_{3}b_{2}c_{3}, a_{2}b_{3}c_{2}a_{3}b_{2}c_{2}a_{3}b_{1}c_{1}+a_{1}b_{3}c_{1}]This is equal to (c�a)b  (b�c)a, which you can verify using the same method as in the Lagrange formula, above.
Some advice:
In both of these identities, you have (c�a)b as a positive number, with the other vector subtracted from this one. Look at them again:
a�(b�c) = (c�a)b  (b�a)c  because the resulting vector is in the bc plane
(a�b)�c = (c�a)b  (b�c)a  because the resulting vector is in the ab planeCall me a cab!
(OK, you're a cab!)
The other Jacobi Identity
(a�b)�c + (b�c)�a + (c�a)�b = 0
Proof:
Add the "CAB Minus BCA" identities:
(a�b)�c = (c�a)b  (b�c)a
(b�c)�a = (a�b)c  (c�a)b
(c�a)�b = (b�c)a  (a�b)c
Three Vectors a, b, and c have a common initial point. Their endpoints form a triangle whose area is given by the magnitude of the vector:
1/2(a�b + b�c + c�a)
(a�b)�((b�c)�(c�a)) = (a�(b�c))^{2}
Triple Product  a·(b�c) is a scalar representing the "signed volume" of a parallelepiped
Triangle Area Using Vectors, part 1 and part 2
Triangle Area using Determinant
Geometry and Trigonometry, and in particular, the Points and Lines section.
The webmaster and author of this Math Help site is Graeme McRae.